Expected value - How many randomly positioned quarter-circles to fill a circle

Question

Suppose that, on top of a circle, you have a circular sector with a central angle of 90° and that the sector can spin about the center, always covering 1/4 of the circle below.  Suppose you spin the sector, let it stop, and color in the area of the circle where the sector stopped.  How many times would you expect to spin the sector in order to color in the entire circle?

Can this be generalized for a central angle of m°?

I have not been sure how to approach this problem.  At one point, I simply tried to find what percent of the circle you'd expect to be colored in after 2 spins, 3 spins. However, I realized that this is not an equivalent question, though perhaps this method might yield some means of approximation. 

At another point, I began thinking about the problem as in some ways akin to a coupon collector problem with coupons collected in groups So, imagine the circle has 360 sectors of 1° each. You spin the 90° sector then fill in or "collect" the ninety 1° sectors where the spinner stopped until you have filled in or "collected" all 360. I don't know how to work out this sort of variant on the coupon collector problem, which is different than the variant in the link: in the scenario I've described, you collect 90 contiguous sectors per spin, not just any 90. If it is possible to adapt the coupon collector approach in the way described, this may provide a means of approximation as well, though the original problem posed at the top does not have a finite number of items to collect or a finite number of positions where the spinner may stop.

This is where I ran out of ideas or know-how for further steps or a different method.

Thank you very much for your help and insights!

Answer 1

To find the probability that the entire circle is coloured after $n$ spins, consider the $n$ angles at which one of the edges of the sector (say, the counterclockwise one) stopped. There’s a gap in the colour if there’s a gap of at least $\frac\pi2$ between any two of these angles. For any given angle, the probability that none of the other $n-1$ angles is within $\frac\pi2$ clockwise of it is $\left(\frac34\right)^{n-1}$. If these events were mutually exclusive, we could add their probabilities to get a probability $n\left(\frac34\right)^{n-1}$ for there to be a gap. But they are not mutually exclusive; we’d be double-counting configurations with two gaps. And there is even room for three gaps. So we need to apply inclusion–exclusion.

There are $\binom nk$ ways to select $k$ particular angles, and the probability for there to be a gap clockwise of each of these $k$ angles is $\left(1-\frac k4\right)^{n-1}$. (Fix one angle; choose the $n-1$ intervals that determine the other $n-1$ angles so they add up to $\left(1-\frac k4\right)2\pi$; then insert $k$ additional gaps of length $\frac\pi2$ at the chosen locations.) Thus, by inclusion–exclusion the probability for there to be at least one gap of length at least $\frac\pi2$ is

$$ \sum_{k=1}^3(-1)^{k+1}\binom nk\left(1-\frac k4\right)^{n-1}\;. $$

This result is also proved (for general sector length) as Theorem $2$ in Section I.$9$ on p. $28$ of William Feller’s An Introduction to Probability Theory and Its Applications ($2$nd edition).

For the expected value of the number $N$ of spins needed to colour the entire circle, this yields

\begin{eqnarray} \mathsf E[N] &=& \sum_{n=0}^\infty\mathsf P(N\gt n) \\ &=& 1+\sum_{n=1}^\infty\sum_{k=1}^3(-1)^{k+1}\binom nk\left(1-\frac k4\right)^{n-1}\;. \\ &=& 1+\sum_{k=1}^3(-1)^{k+1}\sum_{n=1}^\infty\binom nk\left(1-\frac k4\right)^{n-1}\;. \\ &=& 1+\sum_{k=1}^3(-1)^{k+1}\left(1-\frac k4\right)^{k-1}\left(\frac4k\right)^{k+1} \\ &=& 1+\sum_{k=1}^3(-1)^{k+1}\sum_{n=1}^\infty\binom nk\left(1-\frac k4\right)^{n-1}\;. \\ &=& 1+16-4+\frac{16}{81} \\ &=& \frac{1069}{81} \\[5pt] &\approx& 13.2\;. \end{eqnarray}

Here’s Java code that checks this result by simulation.